If we expand the product, we have
[tex]f(x)=-(x+3)(x-14)=-x^2+11x+42[/tex]
The x-coordinate of the vertex is given by
[tex]V_x=-\dfrac{b}{2a}=-\dfrac{11}{-2}=\dfrac{11}{2}[/tex]
Plug this value in the equation to find the correspondant y-value:
[tex]f\left(\dfrac{11}{2}\right)=-\left(\dfrac{11}{2}\right)^2+11\left(\dfrac{11}{2}\right)+42=-\dfrac{121}{4}+\dfrac{121}{2}+42[/tex]
